JOURNAL
OF NUMBER
THEORY
41, 225-230
An Inequality
(1992)
for Multiplicative
Functions
K. SOUNDARARAJAN
28, First Main Road,
Department
qf Mathematics,
Newa Colony, Chromepet.
University
of Michigan,
Communicated
Received
by Hans Zassenhaus
July 3, 1990; revised
February
Let k be a positive integer and f a multiplicative
all primes p. Then, for squarefree
n, we have
This
improves
some recent
results
Madras 600 044, India
Ann Arbor, Michigan
48109
of Alladi,
22, 1991
function
ErdBs,
with 0
and Vaaler.
<f(p) < l/k
for
!? 1992 Academic
Press, Inc.
1. INTRODUCTION
In [l], Alladi, Erdiis, and Vaaler proved the following theorem, which
generalises the well known
;
1 =F+o(l).
dCJ;;
Namely,
they proved
THEOREM
with 0 <f(p)
where o(l)+0
1. Let k be a positive integer and f a multiplicative function
< l/k for all primes p. Then, for squarefree n, we have
as o(n)=&,
1 -+ co.
The object of this paper is to improve and extend Theorem 1. The proof
that we present is also simpler than that of [l].
225
0022-314X/92
$5.00
Copyright
Q 1992 by Academic Press, Inc.
All rights of reproduction
in any form reserved.
226
K.SOUNDARARAJAN
2. NOTATION
Here, we explain the notation used.
Let t B 0 and F, be the set of multiplicative
functions F: { 1,2, 3, ...) --t
[0, oo), which satisfy F(p) > t for all primes p. Also, we denote by G, the
set of all multiplicative
functions G: ( 1, 2, 3, . 3 -+ [0, cc) which satisfy
t 3 G(p) > 0 for all primes p.
For squarefree integers n, we write
b( t, n) = sup {(
z7
Gld))j~G(d))~‘:GEC,J.
Also we write,
A(t) = inf (a( t, m) : m squarefree)
B( t ) = sup (6( t, m) : m squarefree).
3. STATEMENT OF RESULTS
Clearly, for all FE F,, we have
;n
F(434t)
d> &lra 11
1 F(d),
din
where n is squarefree. It is obvious that Theorem 1 is equivalent to saying
A(k)22k+2+o(l)
l
for integers k > 1.
Improving this, we prove
THEOREM 2.
For all t 2 0, we have
A(t+
A(t)
l)>-----A(t)+
In particular
1
k+l
A(k) 2 for non-negative integers k.
Regarding B(t), we prove
1’
MULTIPLICATIVE
THEOREM
FUNCTIONS
227
3. For all t 3 0, we have
1
B(t+ l)<--2-B(t)’
In particular, we have
for positive integers k.
We, use Theorem 3 to extend Theorem 1 for rational numbers k 2 0. WP
prove
THEOREM
4. For all t > 0, we have
A( l/t) + B(t) = 1.
Further, $ k >O is rational and [a,, a,, .... a,] is the continued fraction
expansion of k, then we have
A(k) >,
1
l+a,+a,+
... +a,
and
B(k) <
a, + a, + . . . + a,
1 +a,+a,
+ ... +a,’
4. PROOF OF THEOREM 2
Let F be any element of Fz+, (t > 0). We write F(n) = Cdln g(d) where g
is a multiplicative
function. Clearly g E F,. Now
1
f’(d)= dinc
c g(b)
ub = d
d>,,(‘+I)/(f+2)
=c
aln
g(b)
c
blnla
bg(,+‘+i)/(l+2)/,,
c
641141/Z-8
g(b)
228
and so A(t+ l)aA(t)/(A(t)+
1). Now clearly, A(O)= 1 and hence it
follows that A(k) B l/(k + 1) for all non-negative integers k. This proves
Theorem 2.
5. PROOF OF THEOREM
3
Let G be any element of G,, 1 (t > 0). As before, we write
G(n) = Cdln h(d) where h is multiplicative
and clearly belongs to G,. Now
1
=c4n bin
c h(b)+uln
c h(b)
blnla
o<,,1/11+2)
b,(,,i+!)/fl+~lja)
u~nliti+2)
b>(,,it+i)/(I++)
<B(t)
1
GWl+
u/n
o<nv(l+21
1
W/a)
alla
.>“vlt+2l
(by the definition of B(t))
<I
1
G(a)+ c G(u)oln
a>nIr+llilf+zl Qln
1
G(a)
a/n
a>“lr+~M~r+21
and hence B( t + 1) < l/(2 - B(t)). Clearly B( 1) = l/2, and so B(k) <
k/(k + 1) for all positive integers k.
For a fixed positive integer k, consider a squarefree number n having a
prime factor, p, greater than n kl(k + ‘), Consider the function G’(d) = kocd)
which belongs to Gk. Clearly
c
din
d 2 &lx
km(d)=
c
k+)+l
=k
pdln
+ 11
=
k(k
C
dlnlp
+
1 )“‘(“I
- ’
kw(d)
MULTIPLICATIVE
229
FUNCTIONS
and
whence
1 kwcd)= (k + 1 )w(d),
din
B(k) 2 h(k, n) 3 k/(k + 1).
From
this and our earlier statement B(k) d k/(k + 1) it follows
B(k) = k/(k + l), which completes the proof of Theorem 3.
that
6. PROOF OF THEOREM 4
Let G be an element of G, and let H(I) = l/(G(l))
Then HE F(,,,, (t > 0). Consider,
for all squarefree I.
From this, it easily follows that
A( l/t) + B(t) = 1
for
t>O.
To prove the second part of the theorem, it sufhces to consider k in
(0, 11, by Theorems 2 and 3. We use induction on the denominator of k.
The theorem is easily seen to be true for all rational numbers with
denominator 1. Suppose the theorem is true for all rational numbers with
denominator less than the denominator of k. Let [0, ur, .... a,] be the continued fraction expansion of k E (0, l] and [a,, .... a,] that of l/k. Now l/k
has a lower denominator than k, and the result follows from the induction
hypothesis and the fact
A(k)+B(l/k)=B(k)+A(l/k)=l.
This proves the theorem.
230
K. SOUNDARARAJAN
7. CONCLUDING
REMARKS
If t is an irrational number, then by considering a rational approximawe
tion a/q to t from above, with q’6 m o n and O<a/q-td2/(o(n)),
can see (using Theorem 4) that
1
10&i?
a(& n) B
and
b( t, n) d 1 -
1
10&j’
This improves Theorem 2 of [ 1 ] which gives a(t, n) > t/((t + 1) w(n)).
However, in view of Theorem 4, we would expect bounds for a(?, n) and
b(t, n) which are independent of w(n).
ACKNOWLEDGMENTS
The author is indebted to Professor R. Balasubramanian for suggesting the topic in the first
place and for encouragement. The author is also extremely grateful to the referee for his
valuable suggestions.
REFERENCE
1. K. ALLADI,
J. Number
P. ERD~S,
Theory
AND
J. D. VAALER,
31 (1989). 183-190.
Multiplicative
functions and small divisors, II,